Dividing Radicals

Date: 02/15/99 at 17:34:33
From: Doctor Pat
Subject: Re: Dividing Radicals
To divide by a simple radical like the one in your problem, simply
multiply both the numerator and denominator by the radical. This is
called "rationalizing the denominator" and will leave a whole number in
the denominator. Here is what it looks like:
7 sqrt32 7 sqrt(32)*sqrt(63) 7 sqrt(32*63)
-------- = ------------------- = -------------- =
5 sqrt63 5 sqrt(63)*sqrt(63) 5 * 63
Now all you have to do is simplify. The 7 in the numerator will divide
into the 63 in the denominator, and the sqrt (32*63) can be simplified
also (HINT: look for perfect square factors).
- Doctor Pat, The Math Forum
http://mathforum.org/dr.math/

Date: 02/17/99 at 10:50:19
From: Doctor Bonnie
Subject: Re: Dividing Radicals
Let us first break down all the stuff that we are taking the square
root of into primes:
7 sqrt(2*2*2*2*2)
-----------------
5 sqrt (3*3*7)
So now we can do some actual work. Group the 2's in the numerator into
groups and the 3's in the denominator:
7 sqrt{(2*2)*(2*2)*2}
---------------------
5 sqrt{(3*3)*7}
Because all the stuff under the sqrt sign is multiplied, we can take
the sqrt of them separately:
7 sqrt(2*2) sqrt(2*2) sqrt(2)
-----------------------------
5 sqrt(3*3) sqrt(7)
So now we can take some sqrts:
7*2*2 sqrt 2
------------
5*3 sqrt 7
All we can do now is rationalize the denominator. This is not
necessary, but it is sometimes fun to do, and it might lead to a bit
simpler answer:
(7*2*2 sqrt2)*(sqrt7)
---------------------
(5*3 sqrt7)*(sqrt7)
So we get:
7*2*2 sqrt 14
-------------
5*3*7
And doing the multiplication and canceling:
4 sqrt 14
---------
15
So I think this is about as pretty as you are going to get it.
How about you try some?
3 sqrt 8
----------
11 sqrt 44
and:
2 sqrt 18
---------
3 sqrt 12
- Doctor Bonnie, The Math Forum
http://mathforum.org/dr.math/